Chapter 15, Problem 53A

Interpretation Introduction

Interpretation:The mass of silver chloride formed, and the concentration of silver ions present in the solution has to be calculated.

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 53A

The mass of silver chloride formed is $\text{0}\text{.0717}\text{g}\text{.}$

The concentration of silver ions present in the solution is $\text{1}\text{.17}\text{M}\text{.}$

Explanation of Solution

The chemical reaction is written as,

${\text{AgNO}}_3+\text{NaCl}\rightleftharpoons \text{AgCl}+{\text{NaNO}}_{\text{3}}$

Given,

Volume of sodium chloride is $\text{50}\text{mL}\text{.}$

Molarity of sodium chloride is $\text{1}{\text{.0×10}}^{\text{-2}}\text{M}\text{.}$

Mass of silver nitrate is $\text{10}\text{.0}\text{g}\text{.}$

Molarity: Molarity is defined as the number of moles of solute in one liter of solution.The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

The moles of silver nitrate is calculated as,

Moles= $\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Moles= $\frac{\text{10}\text{.0}\text{g}}{\text{169}\text{.87}\text{g/mol}}$

Moles= $\text{0}\text{.059}\text{mol}$

The moles of silver nitrate is $\text{0}\text{.059}\text{mol}\text{.}$

The moles of sodium chloride is calculated as,

Moles of sodium chloride= $\text{Molarity×Volume}$

Moles of sodium chloride= $\text{1}{\text{.0×10}}^{\text{-2}}\text{mol×0}\text{.05}\text{L}$

Moles of sodium chloride= $\text{0}\text{.0005}\text{mol}$

The moles of sodium chloride is $\text{0}\text{.0005}\text{mol}\text{.}$

One mole of sodium chloride requires one mole of silver nitrate.

Thus, $\text{0}\text{.0005}\text{mol}$ of sodium requires $\frac{\text{1}}{\text{1}}\text{×0}\text{.0005}\text{mol=0}\text{.0005}\text{mol}{\text{AgNO}}_{\text{3}}$

Sodium chloride is the limiting reagent as it limits the formation of product and silver chloride is the excess reagent.

The mass of silver chloride is calculated as,

Mass of silver chloride= $\text{Mass×Molar}\text{mass}$

Mass of silver chloride= $\text{0}\text{.0005}\text{mol×143}\text{.5}\text{g/mol}$

Mass of silver chloride= $\text{0}\text{.07175}\text{g}$

The mass of silver chloride formed is $\text{0}\text{.0717}\text{g}\text{.}$

The moles of silver ions left is calculated as,

Moles of silver ions (left)=Moles of silver nitrate-Moles of sodium chloride

Moles of silver ions (left)= $\text{0}\text{.059}\text{mol-0}\text{.0005}\text{mol}$

Moles of silver ions (left)= $\text{0}\text{.0585}\text{mol}$

The moles of silver ions left is $\text{0}\text{.0585}\text{mol}\text{.}$

The molarity of silver ion in the solution is calculated as,

  $\begin{array}{l}\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}\\ Molarity=\frac{\text{0}\text{.0585}\text{mol}}{\text{0}\text{.05}\text{L}}\\ Molarity=\text{1}\text{.17}\text{M}\end{array}$

The concentration of silver ions present in the solution is $\text{1}\text{.17}\text{M}\text{.}$

Conclusion

The mass of silver chloride formed is $\text{0}\text{.0717}\text{g}\text{.}$

The concentration of silver ions present in the solution is $\text{1}\text{.17}\text{M}\text{.}$